Problem 28 - An Introduction to the Mechanics of Solids in SI Units

Question

The steering system of an automobile consists of a steering wheel, of a steering column which is a shaft of \(2 \mathrm{~cm}\) diameter, and of a linkage which gives a \(20: 1\) reduction in angular rotation between the steering wheel and the tires. Each of the front wheels carries \(5 \mathrm{kN}\) of the weight of the car, and the tires are inflated to a pressure of \(200 \mathrm{kPa}\). If the coefficient of friction between rubber and the ground is \(0.6\), calculate the maximum stress set up in the steering column while the wheels are being turned. (Note: To keep the calculation simple, assume that the contacting region between tire and ground is circular, and that the pressure over it is uniformly \(200 \mathrm{kPa}\) ).

Step-by-Step Solution

Answer

The maximum stress set up in the steering column while the wheels are being turned is \(764 MPa\).

Step 1: Calculate the friction force

The force exerted by the tire due to friction can be calculated using: \[ F = \mu \times W \]where:\( F \) is the friction force,\( \mu \) is the coefficient of friction, and\( W \) is the weight of the car on each tire, which is equal to 5kN or \( 5000 N \).Substituting the known values we get:\[ F = 0.6 \times 5000 = 3000 N \]

Step 2: Determine the Torque

This steering system has a 20:1 reduction in the angular rotation. Thus, the torque needed to keep a rotation can be calculated as:\[ T = 20 \times F \]Substituting the above calculated value for F we get:\[ T = 20 \times 3000 = 60000 N \cdot m \]

Step 3: Calculate Maximum Stress

The maximum stress caused by torsion in a circular shaft is given by:\[ \sigma = \frac{T \cdot r}{J} \]Where:\( \sigma \) is the maximum stress,\( T \) is the torque,\( r \) is the radius of the shaft, and\( J \) is the polar moment of inertia for a circular cross section.First, we should calculate the radius and moment of inertia. For this, we know that the diameter is 2cm or 0.02m, therefore the radius \( r \) is \( \frac{d}{2} = \frac{0.02}{2} = 0.01m \).The polar moment of inertia for a circular cross section is given by \( J = \frac{\pi r^4}{2} \), which gives \[ J = \frac{\pi (0.01)^4}{2} = 7.85 \times 10^{-9} \, m^4 \].Substituting these values into the formula for maximum stress, we get:\[ \sigma = \frac{60000 \cdot 0.01}{7.85 \times 10^{-9}} = 7.64 \times 10^{8} Pa \]or\[ \sigma = 764 MPa \]

Key Concepts

Calculating Friction Force

Understanding how the friction force is calculated is crucial in many fields of engineering, including the automotive industry. The friction force is that resisting force encountered when two surfaces slide against each other. It is determined by the formula:
\[ F = \text{\(\mu\)} \times W \]
where \( F \) represents the friction force, \( \mu \) is the coefficient of friction between the two surfaces, and \( W \) stands for the normal force, usually due to the object's weight. In the case of a car's tire on the road, \( W \) would be the car's weight on the tire, and for a car with each tire bearing 5kN or \( 5000 N \), the friction force can be substantial. This force plays a significant role in ensuring the vehicle's traction as it moves or turns, and engineers must calculate it to design efficient and safe steering systems.
When the coefficient of friction (\( \mu \)) and weight (\( W \)) are known, the calculation is straightforward. For example, with a \( \mu \) of 0.6 for rubber on the road and \( W \) of 5000 N (5 kN), the friction force (\( F \)) would be \( 3000 N \), which has a direct impact on the subsequent calculations of torque and stress within the steering mechanism.

Torque in Steering Systems

In a steering system, torque is that turning force applied to steer the wheels. It's crucial for maneuvering the vehicle and overcoming the resistance due to friction between the tires and the road surface. The magnitude of torque required in a steering system depends on several factors including the friction force and the mechanical design of the steering linkage.
In the provided example, the system is designed with a 20:1 reduction ratio, meaning that the torque applied on the steering wheel is reduced by 20 times when it reaches the tires. This mechanical advantage allows for easier turning of the wheels by the driver. Now, if the friction force (\( F \)) is \( 3000 N \), the torque (\( T \)), required to turn the wheels can be deduced by multiplying the friction force by the reduction ratio:
\[ T = 20 \times F = 20 \times 3000 = 60000 N \cdot m \]
This calculated torque is essential for determining the stress experienced by the steering column, which must be well within the safety limits to prevent mechanical failure.

Calculating Maximum Stress in a Shaft

Maximum stress in a shaft is a critical factor in the design and integrity of various mechanical systems. Stress within a shaft is directly related to the applied torque and the shaft's physical characteristics. For a circular shaft, the formula to calculate the maximum stress (\( \sigma \)) caused by the applied torque (\( T \)) is:
\[ \sigma = \frac{T \cdot r}{J} \]
Here, \( r \) is the radius of the shaft and \( J \) is its polar moment of inertia, which depends on the shaft's geometry. In the context of a steering column with a 2 cm diameter, you calculate the radius by halving the diameter (0.02 m / 2 = 0.01 m) and the polar moment of inertia for a circular cross-section with \[ J = \frac{\pi r^4}{2} \].
Using these values with the previously found torque of \( 60000 N\cdot m \), you would get a substantial stress of 764 MPa, a necessary calculation for ensuring the shaft is designed to withstand the exerted forces without failing. This calculated stress must be less than the material's yield strength to avoid permanent deformation or other failures, securing the reliability and safety of the steering system.